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The continued-fraction algorithm terminates exactly on rational numbers

Statement

For a real number $α$ , the continued-fraction algorithm terminates if and only if $α$ is rational. When $α$ is rational, the continued-fraction digits are exactly the successive quotient digits of the Euclidean algorithm.

Facts & Assumptions

Given: A real number $α$ and its continued-fraction algorithm.

[F1]

If $u, v \in Z$ with $v > 0$ , then there are unique integers $a, r$ with $u = a v + r, 0 \leq r < v .$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[F2]

A terminating continued-fraction expansion is a finite regular continued fraction, and every finite regular continued fraction equals its last convergent $p_{n} / q_{n} \in Q$ (The continued-fraction algorithm for real numbers, Convergents are given by the standard recurrences and tail formula).

[F3]

Every complete quotient after the first is greater than $1$ (Complete quotients in the continued-fraction algorithm).

Proof

technique · direct

1.1

If $α$ is an integer, the algorithm stops immediately. Otherwise write $α = u_{0} / v_{0}$ with integers $v_{0} > 0$ and $\gcd (u_{0}, v_{0}) = 1$ . Applying [F1] to. [given, F1, algebra] $u_{0}$ and $v_{0}$ gives $u_{0} = a_{0} v_{0} + r_{0}, 0 < r_{0} < v_{0},$ so $α = a_{0} + \frac{r_{0}}{v_{0}}, α_{1} = \frac{1}{α - a_{0}} = \frac{v_{0}}{r_{0}} .$ Thus the next numerator is the previous denominator.

givenF1algebra

1.2

Conversely, if the algorithm terminates, then by [F2] the original number is a finite regular continued fraction and hence rational.

1.3

Let $n \geq 1$ , and suppose. [F1, F3, induction, algebra] $α_{n} = \frac{u_{n}}{v_{n}} (v_{n} > 0, \gcd (u_{n}, v_{n}) = 1)$ is not an integer. Since $n \geq 1$ , fact [F3] gives $α_{n} > 1$ , so $u_{n} > v_{n} > 0$ . Applying [F1] gives $u_{n} = a_{n} v_{n} + r_{n}, 0 < r_{n} < v_{n}, α_{n + 1} = \frac{v_{n}}{r_{n}} .$ Any common divisor of $v_{n}$ and $r_{n} = u_{n} - a_{n} v_{n}$ also divides $u_{n}$ , so $\gcd (v_{n}, r_{n}) = 1$ . Hence this is already the reduced form $α_{n + 1} = \frac{u_{n + 1}}{v_{n + 1}} = \frac{v_{n}}{r_{n}},$ and therefore $u_{n + 1} = v_{n} < u_{n}$ .

F1F3inductionalgebra

2.1

When $α$ is a nonintegral rational, step 1.1 gives $α_{1} = \frac{v_{0}}{r_{0}} > 1.$ So step 1.3 applies successively to $α_{1}, α_{2}, \dots$ as long as they remain nonintegral. Their positive integer numerators $u_{1}, u_{2}, u_{3}, \dots$ then form a strictly decreasing sequence, which cannot continue forever. Therefore some complete quotient is an integer, and the algorithm terminates.

step 1.1step 1.3induction

3.1

The equations in steps 1.1 and 1.3 are exactly the Euclidean divisions of the successive numerator-denominator pairs, so the continued-fraction digits are. [step 1.1, step 1.3, F1] the Euclidean quotient digits.

step 1.1step 1.3F1∎

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Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
William Stein, Elementary Number Theory: Primes, Congruences, and Secrets (standard reference, not scraped)